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Question 428471: I need to use substitution method
15x-10y=1
5y=-2+15x
Found 3 solutions by mananth, ikleyn, greenestamps:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! 15 x + -10 y = 1 .............1
5 y = -2 15 x ---------2
/ 5 =
y = -0.4 3 x
Plug the value of y in (1)
15x-10(-0.4+3x)
15x+4-30x=1
15x-30x=1-4
-15x=-3
/-15
x=0.2
Plug the value of x in (1)
15x-10y=1
15*0.2-10y=1
3-10y=1
-10y=-2
/-10
y=0.2
Answer by ikleyn(53419)  (Show Source):
You can put this solution on YOUR website! .
I need to use substitution method
15x - 10y = 1
5y = -2 + 15x
~~~~~~~~~~~~~~~~~~~~~~~~
It can be solved in much more efficient way than @mananth does it in his post.
I will apply a " smart " substitution method.
Your starting equations are
15x - 10y = 1 (1)
5y = -2 + 15x (2)
From equation (2), express 15x = 5y + 2 and substitute it as a whole block into equation (1),
replacing the term 15x there
(5y + 2) - 10y = 1,
5y - 10y = 1 - 2,
-5y = -1
y = 1/5 = 0.2.
Now substitute y = 0.2 into equation (1) and find x
15x - 10*0.2 = 1,
15x - 2 = 1,
15x = 1 + 2 = 3,
x = 3/15 = 0.2.
ANSWER. x = 0.2, y = 0.2.
Solved.
As you see, in this case " smart " substitution works more effectively
than a standard " stupid " substitution.
The meaning of this problem for you is to learn advantages of "smart" substitution method.
Answer by greenestamps(13258)  (Show Source):
You can put this solution on YOUR website!
Pay close attention to the solution from tutor @ikleyn.
In a typical algebra class, given a system of two linear equations, you are taught to solve one of the equations for one of the variables (by itself) and substitute the result in the other equation.
But as the tutor shows in this problem, the solution is much easier because the expression "15x" appears in both of the given equations; so solving one of the equations for "15x" and substituting in the other equation makes the solution easy.
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